googologywikiaorg-20200223-history
User blog:PlantStarAlpineer0/Introducing Lightspeed Slash Notation
I’d like to introduce perhaps one of the fastest notations ever invented, called “Lightspeed Slash Notation”. It adopts Chris Bird’s notations to create a new, fast notation. Here are some rules: Suppose that @1 is a string of n-1 1’s and @2 is a string of n-1 2’s: {10,10[@12/n2@2]2} = /n /{10,10[@12/n2@2]2} = /0,n /a,{10,10[@12/n2@2]2} = /a+1,n /{10,10[@12/n2@2]2},0 = /0,0,n There is a pattern here. Suppose that @3 is any string of terms: /@3,{10,10[@12/n2@2]2} = /@3+1,n where 1 is added to the last term in the string if the term is not {10,10@12/n2@2]2} Here is where we hit another ceiling. No matter, we can come up with another delimiter with further rules. Let’s define the “letter A” delimiter as coming after all numbers that obey the above rules. Let’s also suppose that @4 is any string of 0’s that is {10,10[@12/n2@2]2} terms long: /@4 = /An /A@4 = /Bn /B@4 = /Cn Here we have another pattern. Suppose that @L is any letter in the English alphabet: /@L@4 = /(@L+A)n But what happens if we hit the end of the letter Z? Well, let’s suppose that we can add letters like we would with basic addition: /Z+@L@4 = /Z+(@L+A)n From here, let’s say we can multiply, exponentiate, tetrate, etc. these letters. Suppose that @L2 is any string of Z’s and b is a letter corresponding to the number of Z’s in both strings: /@L2+@L2n = /Zbn /@L2@L2n = /Zbn Suppose that @O is the minimum number of Z’s needed to reach a successive operation and # is the current operation in the set: /@O#n = /Z(#-2)bn where #-2 translates to the amount of ^'s in the equation, if any What’s next? Let’s define the “double slash” delimiter as coming after all operations that obey the above rules. Let’s also suppose that @L3 is any string of letters in any operation that has {10,10[@12/a2@2]2} ^’s: /@L3n = //n Note that all above rules apply to multiple-slashed numbers: //@L3n = ///n ///@L3n = ////n Here is yet another pattern. Suppose that @/ is any string of slashes: @/@L3n = @//n If there are {10,10[@12/a2@2]2} slashes, we can begin using the aforementioned operations on these slashes. Suppose that @/2 is any string of {10,10[@12/a2@2]2} slashes and d is the last exponent of the result: @/2@/2n = /dn Suppose that @O/ is the minimum number of slashes needed to reach a successive operation: @O/#n = /(#-2)dn where #-2 translates to the amount of ^'s in the equation, if any We have reached yet another ceiling. However, there is an impasse. If the amount of ^’s is equal to {10,10[@12/n2@2]2}: @O/#0 = /0n This way, we can travel ever faster towards the edge of the mathematical omniverse: @O/#a0 = /a+1n This is a good opportunity for nesting. Suppose that @e is a bracketed number with e nests. If the number includes {10,10[@12/n2@2]2} nests: /@e0 = /0,n0 /a,@e0 = /a+1,n0 You get the idea. From here, we’re going meta, until we surpass /Z^^^…^^^B00 with {10,10[@12/n2@2]2} ^’s. We have already used the slash, so we’ll have to resort to using the | symbol. After that, we can use the above four rules within the brackets. But once we’ve exhausted that… well, we can keep coming up with new delimiters and rules. Suppose that @| is any string of {10,10[@12/n2@2]2} ^’s: /[|@||0]0 = /[ 00]n Note that the above rules can apply on a much grander scale here. Suppose that @e2 is any bracketed number in this style with {10,10[@12/n2@2]2} nests: /@e20 = /[ 0 ]n (The space is added to remove links. In reality, the brackets would be next to each other.) Now we can do this: /[ a]@e20 = /[ a+1]n Of course, we can have /@e20 and /[ [ a]]0. But after that? Suppose that @|| is the amount of {10,10[@12/n2@2]2}- Okay, enough calling it that, let’s give it a simpler name: RH. Suppose that @|| is the amount of RH nests in this style: /@||0 = /000n /0a0@L3 = /0a+10n /0@L3a0 = /00a+1n /a0|0 = /a+100n /|000 = /00000n We can use this method with three or more sets of brackets with numbers between them. Once we have exhausted that, we need another delimiter. Suppose that @is the number of such sets in this system. If @[s is equal to RH: /@[s0 = ~/n where @[s is equal to RH These rules can be found on this page: https://plantstarslargenumbers.wordpress.com/m3/ [[Category:Blog posts]